The common form of objection is given in the reply, that so far from paying an infinite sum, no sensible man would give anything approaching to £50 for such a chance. Probably not, because no man would see enough of the series to make it worth his while. What most persons form their practical opinion upon, is such small portions of the series as they have actually seen or can reasonably expect. Now in any such portion, say one which embraces 100 turns, the longest succession of heads would not amount on the average to more than seven or eight. This is observed, but it is forgotten that the formula which produced these, would, if it had greater scope, keep on producing better and better ones without any limit. Hence it arises that some persons are perplexed, because the conduct they would adopt, in reference to the curtailed portion of the series which they are practically likely to meet with, does not find its justification in inferences which are necessarily based upon the series in the completeness of its infinitude. 16. This will be more clearly seen by considering the various possibilities, and the scope required in order to exhaust them, when we confine ourselves to a _limited_ number of throws. Begin with three. This yields eight equally likely possibilities. In four of these cases the thrower starts with tail and therefore loses: in two he gains a single point (i.e. £1); in one he gains two points, and in one he gains four points. Hence his total gain being eight pounds achieved in four different contingencies, his average gain would be two pounds. Now suppose he be allowed to go as far as n throws, so that we have to contemplate 2^{n} possibilities. All of these have to be taken into account if we wish to consider what happens on the average. It will readily be seen that, when all the possible cases have been reckoned once, his total gain will be (reckoned in pounds), 2^{n-2} + 2^{n-3}·2 + 2^{n-4}·2^{2} + ... + 2·2^{n-3} + 2^{n-2} + 2^{n-1}, viz. (n + 1) 2^{n-2}. This being spread over 2^{n-1} different occasions of gain his average gain will be 1/2 (n + 1).